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kifflom [539]
3 years ago
8

what is the sum of a 7-term geometric series if the first term is −6, the last term is −24,576, and the common ratio is −4? −32,

766 −19,662 19,662 32,766
Mathematics
1 answer:
Lena [83]3 years ago
7 0
Sum of ggeo metric sequence is

Sn=\frac{a1(1-r^n)}{1-r}

n=which term
a1=first term
r=common ratio


so
n=7
a1=-6
r=-4

Sn=\frac{-6(1-(-4)^7)}{1-(-4)}
Sn=\frac{-6(1-(-16384))}{1+4}
Sn=\frac{-6(1+16384)}{5}
Sn=\frac{-6(16385)}{5}
Sn=\frac{-98310}{5}
Sn=-19662

answer is 2nd option os -19662



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\bf x^2+15x+\boxed{?}^2\implies \stackrel{\textit{we know the middle term is}}{2\sqrt{x^2}\cdot \sqrt{\boxed{?}^2}\implies 2x\boxed{?}}\qquad then\qquad 2x\boxed{?}=15x \\\\\\ \boxed{?}=\cfrac{15x}{2x}\implies \boxed{?}=\cfrac{15}{2}\qquad \impliedby \textit{we should add that much \underline{squared}} \\\\[-0.35em] ~\dotfill\\\\ x^2+15x+\left( \cfrac{15}{2} \right)^2\implies \left(x+ \cfrac{15}{2} \right)^2

5 0
3 years ago
In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass bala
Maslowich

Answer:

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen        A               B           d = a - b         d²

1                     13.76        13.74         0.02           0.004

2                    12.47        12.45          0.02         0.004

3                    10.09        10.08           0.01        0.001

4                       8.91       8.92          -0.01          0.001

5                     13.57      13.54           0.03        0.009

<u>6                     12.74      12.75          -0.01        0.001</u>

<u>∑                                                      0.06         0.0173</u>

d`= ∑d/n= 0.006/6= 0.001

sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882

sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

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3 years ago
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